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A proton, a deuteron (nucleus of ${ }_1 \mathrm{H}^2$ ) and an $\alpha$-particle with same kinetic energy enter a region of uniform magnetic field moving at right angles to the field. The ratio of the radii of their circular paths is :
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The correct answer is:
$1: \sqrt{2}: 1$
$r=\frac{m v}{q B}=\frac{\sqrt{2 m K}}{q B}$
$\Rightarrow \quad r \propto \frac{(m)^{1 / 2}}{q}$
$\therefore \quad r_p=\frac{\left(m_p\right)^{1 / 2}}{q_p}, r_d=\frac{\left(m_d\right)^{1 / 2}}{q_d}$
and $\quad r_\alpha=\frac{\left(m_\alpha\right)^{1 / 2}}{q_\alpha}$
$\therefore \quad r_p: r_d: r_\alpha=\frac{\sqrt{m}}{q}: \frac{\sqrt{2 m}}{q}: \frac{\sqrt{4 m}}{2 q}$
$=1: \sqrt{2}: 1$
$\Rightarrow \quad r \propto \frac{(m)^{1 / 2}}{q}$
$\therefore \quad r_p=\frac{\left(m_p\right)^{1 / 2}}{q_p}, r_d=\frac{\left(m_d\right)^{1 / 2}}{q_d}$
and $\quad r_\alpha=\frac{\left(m_\alpha\right)^{1 / 2}}{q_\alpha}$
$\therefore \quad r_p: r_d: r_\alpha=\frac{\sqrt{m}}{q}: \frac{\sqrt{2 m}}{q}: \frac{\sqrt{4 m}}{2 q}$
$=1: \sqrt{2}: 1$
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